Enthalpy of reaction

IB Chemistry SL· 5 min read

1. Definition and Sign Conventions★★☆☆☆⏱ 10 min

📘 Definition

Enthalpy of Reaction

\Delta H

The total change in enthalpy (heat content at constant pressure) when a reaction proceeds according to its given balanced stoichiometric equation.

Example:

For (2H_2(g) + O_2(g) \rightarrow 2H_2O(l)), ΔH is the enthalpy change for 2 moles of H₂ reacting with 1 mole of O₂.

The sign of ΔH always describes the change in enthalpy of the chemical system (not the surroundings). The IB uses a consistent convention that is frequently tested in multiple choice questions.

Exothermic reactions: Release heat to the surroundings, so the system's enthalpy decreases. $Δ H < 0$ (negative).
Endothermic reactions: Absorb heat from the surroundings, so the system's enthalpy increases. $Δ H > 0$ (positive).

📐 Worked Example

State the sign of ΔH for the melting of ice to liquid water, and classify the process as exothermic or endothermic.

1
Melting requires heat energy to be absorbed from the surroundings to break the intermolecular forces in ice.
2
The heat is absorbed by the system (the water), so the enthalpy of the system increases.
3
Conclusion: ΔH is positive, and melting is endothermic.

2. Calculation from Bond Enthalpies★★★☆☆⏱ 15 min

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Breaking covalent bonds always requires energy (endothermic, positive energy contribution), and forming covalent bonds always releases energy (exothermic, negative energy contribution). We use this to calculate ΔH for reactions with all gaseous species:

\Delta H = \sum \text{(bond enthalpies of bonds broken)} - \sum \text{(bond enthalpies of bonds formed)}

📐 Worked Example

Calculate ΔH for the reaction: $H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$ . Bond enthalpies (kJ mol⁻¹): H-H = 436, Cl-Cl = 242, H-Cl = 431.

1
Count all bonds broken in reactants: 1 H-H bond and 1 Cl-Cl bond.
2
Sum of bonds broken: $436 + 242 = 678$ kJ mol⁻¹
3
Count all bonds formed in products: 2 H-Cl bonds.
4
Sum of bonds formed: $2 \times 431 = 862$ kJ mol⁻¹
5
Apply formula: $Δ H = 678 - 862 = - 184$ kJ mol⁻¹
6
Negative sign confirms the reaction is exothermic, which matches experimental data.

✓ Quick check

Check your understanding

Which of the following is the correct formula for ΔH from bond enthalpies?
- ΔH = bonds formed - bonds broken
- ΔH = bonds broken - bonds formed
- ΔH = bonds broken + bonds formed
- ΔH = -(bonds broken + bonds formed)
Reveal answer
1 —
Correct! Breaking bonds requires energy (positive input) and forming bonds releases energy (negative output), so ΔH equals total bonds broken minus total bonds formed.

3. Calculation from Standard Enthalpies of Formation★★★☆☆⏱ 15 min

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The standard enthalpy of formation of any element in its standard state is zero. This method can be used for reactions with any state (solid, liquid, gas) and gives the exact standard enthalpy of reaction, unlike average bond enthalpies.

\Delta H^\ominus_{\text{rxn}} = \sum \Delta H_f^\ominus (\text{products}) - \sum \Delta H_f^\ominus (\text{reactants})

📘 Definition

Standard Enthalpy of Reaction

$Δ H_{r x n}^{⊖}$

Enthalpy change for a reaction when all reactants and products are in their standard states at 298 K and 100 kPa.

📐 Worked Example

Calculate $Δ H_{r x n}^{⊖}$ for: $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$ . Given $Δ H_{f}^{⊖}$ (kJ mol⁻¹): $C_{3} H_{8} (g) = - 104$ , $C O_{2} (g) = - 394$ , $H_{2} O (l) = - 286$ , $O_{2} (g) = 0$ .

1
Calculate sum of $Δ H_{f}^{⊖}$ for products: $(3 \times - 394) + (4 \times - 286) = - 1182 - 1144 = - 2326$ kJ mol⁻¹
2
Calculate sum of $Δ H_{f}^{⊖}$ for reactants: $(1 \times - 104) + (5 \times 0) = - 104$ kJ mol⁻¹
3
Apply formula: $Δ H_{r x n}^{⊖} = (- 2326) - (- 104) = - 2326 + 104 = - 2222$ kJ mol⁻¹
4
Negative value confirms combustion is exothermic, which is expected.

4. Common Pitfalls

Wrong move:

Claiming ΔH is positive for exothermic reactions

Why:

Confusing the perspective of the surroundings vs the chemical system. ΔH always measures change in the system.

Correct move:

Remember: Exothermic = system loses enthalpy, so ΔH is negative. Endothermic = system gains enthalpy, so ΔH is positive.

Wrong move:

Reversing the bond enthalpy formula (ΔH = bonds formed - bonds broken)

Why:

Forgetting that bond formation releases energy, which reduces the total enthalpy change of the system.

Correct move:

Use the mnemonic: Broken minus Formed, ΔH = broken - formed.

Wrong move:

Forgetting to multiply enthalpy values by stoichiometric coefficients

Why:

Enthalpy of formation and bond enthalpy are per mole values, so they must scale with the balanced equation.

Correct move:

Always multiply each value by the number of moles (or number of bonds) given by the reaction stoichiometry.

Wrong move:

Using bond enthalpies for liquid or solid species

Why:

Bond enthalpies are only defined for gaseous molecules, and are average values anyway.

Correct move:

Always use the standard enthalpy of formation method for reactions that include liquid or solid reactants or products.

5. Quick Reference Cheatsheet

Concept	Formula	Key Rule
Enthalpy of Reaction	$Δ H = H_{p r o d u c t s} - H_{r e a c t an t s}$	Exothermic: ΔH < 0; Endothermic: ΔH > 0
Bond Enthalpy Method	$Δ H = \sum (b o n d s b r o k e n) - \sum (b o n d s f or m e d)$	Only for all-gaseous reactions, average values
Enthalpy of Formation Method	$Δ H_{r x n}^{⊖} = \sum Δ H_{f}^{⊖} (p r o d u c t s) - \sum Δ H_{f}^{⊖} (r e a c t an t s)$	$Δ H_{f}^{⊖} = 0$ for elements in standard state

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · Paper 1
Calculate ΔH from bond enthalpies
2023 · Paper 2
ΔH from enthalpies of formation
2021 · Paper 1
Identify sign of ΔH for reaction

Going deeper

syllabusIB Chemistry SL Reactivity 2 SyllabusOfficial topic reference

What's Next

Enthalpy of reaction forms the foundation of all thermochemistry topics in IB SL Chemistry. The sign conventions and calculation methods you learned here are required for Hess's law, which allows you to calculate ΔH for reactions that cannot be measured directly experimentally. You will later apply these concepts to calculate enthalpy changes from experimental calorimetry data, and eventually explore entropy and the spontaneity of chemical reactions. Mastery of this sub-topic is critical for all higher level energetics questions in your final exam.